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Is $\mathbb{Z} [i] $ is field ? yes/No

yes, I thinks it will field because it is integral domain

Is its True ?

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closed as off-topic by The Count, mrtaurho, Shailesh, Leucippus, ronno Jun 29 at 17:00

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    $\begingroup$ Try to divide $1$ by $2$. $\endgroup$ – darij grinberg Jun 29 at 14:37
  • $\begingroup$ im not getting @darijgrinberg $\endgroup$ – jasmine Jun 29 at 14:38
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    $\begingroup$ $1/2 \notin \mathbb{Z}[i]$ but if it were a field, $2$ would have to have a multiplicative inverse. $\endgroup$ – mathphys Jun 29 at 14:38
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    $\begingroup$ $\mathbb Z$ is an integral domain. Is it a field? Why not? $\endgroup$ – lulu Jun 29 at 14:40
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    $\begingroup$ Do you know the definitions of the words, "field", "integral domain"? I ask because you say you think it is a field because it is an integral domain. The set of all integers (with the usual addition and multiplication) is an integral domain (that's where the name comes from) but not a field. (Every field is an integral domain. Not every integral domain is a field.) $\endgroup$ – user247327 Jun 29 at 14:42
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No, it is not a field. For instance, $2$ has no inverse in $\mathbb Z[i]$.

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Well, ${\Bbb Z}[i] = \{a+bi\mid a,b\in{\Bbb Z}\}$ is a subset of ${\Bbb C}=\{a+bi\mid a,b\in{\Bbb R}\}$ and so an integral domain. But its clearly not a field.

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Note that $\mathbb{Z} \subseteq \mathbb{Z}[i]$ and $\mathbb{Z}$ is not a field. To see this, we know that $1$ is the identity in $\mathbb{Z}$, so how would we invert $2$? We would need to multiply it by $\frac{1}{2}$, which is not an element of $\mathbb{Z}$. In fact, the only invertible elements in $\mathbb{Z}$ are $\pm 1$, hence the integers do not form a field.

Hence, $\mathbb{Z}[i]$ is not a field.

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We know that $$\mathbb Z[i]\cong\mathbb Z[x]/(x^2+1).$$

And $$(x^2+1)\subsetneq (3,x^2+1)\subsetneq\mathbb Z[x]\quad \text{(why?)}$$ implies $(x^2+1)$ is not a maximal ideal of $\mathbb Z[x]$, hence $\mathbb Z[i]$ is not a field.

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